Question

A particle of mass $2\text{kg}$ moving in positive x-direction with speed $6\text{m/s}$ is hit by another particle of mass $4\text{kg}$ moving in the positive y-direction with speed $3\text{m/s}$. If the collision is perfectly inelastic, the percentage loss in energy during the collision is close to

• A. 50%
• B. 44%
• C. 62%
• D. 56%

Answer 1

The correct option is D 56%

Given that,
$m_1=2\text{kg}$
$v_1=6\text{m/s}$
$m_2=4\text{kg}$
$v_2=3\text{m/s}$


Initial momentum of the system is
$\overrightarrow{p_i}=m_1{v}_1\hat{i}+m_2{v}_2\hat{j}$
So, $p_i=\sqrt{(m_1{v}_1{)}^2+(m_2{v}_2{)}^2}$
= $\sqrt{(12{)}^2+(12{)}^2}=12\sqrt{2}\text{kg m/s}$

Since collision is perfectly inelastic, so $m_1+m_2$ will be net mass after collsion
Using PCLM, $p_i=p_f$
$\Rightarrow 12\sqrt{2}=6v$
$\Rightarrow v=2\sqrt{2}\text{m/s}$

Now, Loss in energy.
$△E=E_1-E_2$
$=[\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2]-\left[\frac{1}{2}(m_1+m_2)v^2)\right]$
$=(\frac{1}{2}\times 2\times 6^2+\frac{1}{2}\times 4\times 3^2)-(\frac{1}{2}\times 6\times (2\sqrt{2}{)}^2)$
$=(36+18)-24=54-24=30\text{J}$
So, $△E_y=\frac{△E}{E_1}\times 100=\frac{30}{54}\times 100=55.55\approx 56\%$